Question 1
A bar magnet at 30° to a 0.25 T field experiences a torque of 4.5×10⁻² J. Find its magnetic moment.
Steps only
m=τ/(B sinθ)=0.045/(0.25×0.5)=0.36 J/T
Answer
0.36 J/T
CLASS 12 · PHYSICS · CHAPTER 5
Magnetism and matter
Simple worked answers to all 7 current exercises on magnets, solenoids and materials.
7 worked answersComplete chapter answers
7 worked answers · short steps first, explanations when you need them
Chapter 5 Exercises
Textbook pages 152–153
Question 2
A magnet of moment 0.32 J/T is in a 0.15 T field. Find the stable and unstable orientations and the potential energy in each.
Steps only
U=−mB cosθ; parallel: −0.048 J (stable); opposite: +0.048 J (unstable).
Answer
Parallel, −0.048 J; opposite, +0.048 J
Question 3
An 800-turn solenoid has cross-sectional area 2.5×10⁻⁴ m² and carries 3.0 A. Explain how it acts like a bar magnet and find its magnetic moment.
Steps only
m=NIA=800(3.0)(2.5×10⁻⁴)=0.60 A m²
Answer
Bar-magnet-like field; m=0.60 A m²
Question 4
For the solenoid described in Exercise 5.3, find the torque in a 0.25 T horizontal field when its axis is at 30° to the field.
Steps only
τ=mB sin30°=(0.60)(0.25)(0.5)=0.075 N m
Answer
0.075 N m
Question 5
A magnet of moment 1.5 J/T starts aligned with a 0.22 T field. Find the external work to turn it (i) perpendicular and (ii) opposite to the field, and the torque in each final position.
Steps only
ΔU: 0.33 J to 90°, 0.66 J to 180°; τ=mB sinθ: 0.33 N m and 0.
Answer
0.33 J, 0.33 N m; 0.66 J, 0
Question 6
A 2000-turn solenoid of area 1.6×10⁻⁴ m² carries 4.0 A. It is free to turn in a horizontal plane. Find (a) its magnetic moment, (b) the force and torque in a uniform 7.5×10⁻² T horizontal field at 30° to its axis.
Steps only
m=NIA=1.28 A m²; Fnet=0; τ=mB sin30°=0.048 N m
Answer
1.28 A m²; zero force; 0.048 N m
Question 7
A short magnet has magnetic moment 0.48 J/T. Find the field at 10 cm from its centre (a) on its axis and (b) on its equatorial line, including direction.
Steps only
Baxis=μ₀(2m)/(4πr³)=9.6×10⁻⁵ T; Beq=μ₀m/(4πr³)=4.8×10⁻⁵ T (opposite m).
Answer
9.6×10⁻⁵ T axial; 4.8×10⁻⁵ T equatorial